What is the foci ((x-4)^2)/(24)+((y-2)^2)/(49)=1 ?

The foci ((x-4)^2)/(24)+((y-2)^2)/(49)=1 is (4,7),(4,-3)

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foci ((x-4)^2)/(24)+((y-2)^2)/(49)=1

foci

\frac{( x - 4 ) ^{2}}{24} + \frac{( y - 2 ) ^{2}}{49} = 1

(4, 7), (4, - 3)

Solution steps

(h, k + c), (h, k - c)

Calculate ellipse properties

\frac{( x - 4 ) ^{2}}{24} + \frac{( y - 2 ) ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (4, 2), b = 7, a = 26

(4, 2 + c), (4, 2 - c)

Compute

c :

c = 7^{2} - (26)^{2} : 5

(4, 2 + 5), (4, 2 - 5)

Refine

(4, 7), (4, - 3)

x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

\frac{( x + 13 ) ^{2}}{400} + \frac{( y + 6 ) ^{2}}{625} = 1

4 x^{2} + y^{2} - 16 x - 2 y - 19 = 0

\frac{x ^{2}}{4} + (y - 1)^{2} = 1

3 x^{2} + 7 y^{2} = 63

What is the foci ((x-4)^2)/(24)+((y-2)^2)/(49)=1 ?
The foci ((x-4)^2)/(24)+((y-2)^2)/(49)=1 is (4,7),(4,-3)